All categories

3 years ago

(PLEASE HELP will give first correct answer brainliest!!) Which option correctly describes glucose catabolism?

Chemistry

2 answers:

Andreas93 [3]3 years ago

5 0

The correct answer is number 1

Novosadov [1.4K]3 years ago

3 0

Th correct answer is number 1

You might be interested in

2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer soluti

lutik1710 [3]

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

5 0

2 years ago

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

sukhopar [10]

Answer:

$\rho _2=0.22g/L$

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

$PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT$

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

$\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}$

It means we can compute the final density as shown below:

$\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}$

Now, we plug in to obtain:

$\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L$

Regards!

8 0

2 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

PSYCHO15rus [73]

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

6 0

2 years ago

19) Which set of coefficients correctly balances this chemical equation?*

Bas_tet [7]

Answer:

6,1,2

Explanation:

_Ag + _N2 => _Ag3N

LHS RHS

Ag=1×6=3 Ag=3×2=6(Balanced)

N=2×1=2 N=1×2=2(Balanced)

coefficient=6,1,2

4 0

2 years ago

I need help name the parts of making up the female part of the flower and state what each part does. ( answer in your own unique

nekit [7.7K]

<h2>Answer:-</h2>

The female part is the pistil. The pistil usually is located in the center of the flower and is made up of three parts: the stigma, style, and ovary. The stigma is the sticky knob at the top of the pistil. It is attached to the long, tube like structure called the style.

6 0

2 years ago